Rev. Anal. Num´er. Th´eor. Approx., vol. 32 (2003) no. 1, pp. 79–84 ictp.acad.ro/jnaat
ON CONVEXITY–LIKE INEQUALITIES (II)
JOSIP PE ˇCARI ´C∗and SANJA VAROˇSANEC†
Abstract. We improve the classical Jensen inequality for convex functions by extending it to a wider class of functions. We also consider some weaker condi- tions for the weights occurring in this inequality.
MSC 2000. 26D15.
Keywords. Jensen’s inequality, convexity-like inequalities, generalized Steffen- sen’s inequality.
1. INTRODUCTION
One of the most important inequality in the theory of convex functions and many fields of mathematics connected with this theory is Jensen’s inequality which states:
Theorem1. If f :I →R(I is an open interval inR) is a convex function, x1, x2, . . . , xn∈I, n≥2, p1, p2, . . . , pn>0, then
(1) fP1
n
Pn i=1
pixi≤ P1
n
n
X
i=1
pif(xi),
where Pn=
n
P
i=1
pi.
This inequality has an important role in mathematics and statistics and there exist a lot of its generalization, improvements and refinements. For ex- ample, it is reasonable to ask whether the conditionpi>0 can be relaxed. Or, on the other hand, can we obtain inequality (8) for a wider class of functions?
Some answers on the second question were given in [1] by Dragomirescu and Ivan, and in [3] by Peˇcari´c and Pearce. Answer on the first question was given by Steffensen [8]. Namely, he obtained that the following theorem holds.
Theorem2. If f :I →R(I is an open interval inR) is a convex function, (x1, x2, . . . , xn)∈In,n≥2,is a monotonicn-tuple, and ifpi are real numbers such that
∗Faculty of Textile Technology, University of Zagreb, Pierrotijeva 6, 10000 Zagreb, Croa- tia, e-mail: [email protected].
†Department of Mathematics, University of Zagreb, Bijeniˇcka 30, 10000 Zagreb, Croatia, e-mail: [email protected].
(2) x¯= P1
n
n
X
i=1
pixi ∈I, 0≤Pk≤Pn, k= 1,2, . . . , n−1, Pn>0,
then(1) holds.
The previous theorem is known in literature under the name: Jensen–
Steffensen’s inequality. In this paper we show that the condition (2) can be relaxed. In Section 2 we suppose that f is a differentiable function and main result is a consequence of the generalized Steffensen’s inequality [7]. In Section 3 we will pointed out that differentiability is not a necessary condition and we will be in position to prove a theorem about Jensen’s type inequality directly, without using some other results. The proof is based on the identity which is first used in the proof of the Jensen-Steffensen inequality by J. Peˇcari´c in [4].
2. CONSEQUENCE OF THE GENERALIZED STEFFENSEN INEQUALITY
We need the following Steffensen type inequality which is given in [7].
Here, we shall use these notation:
t= (t1, . . . , tn), [a,b] =t:ai ≤ti ≤bi,1≤i≤n =
n
Y
i=1
[ai, bi].
If t and x are two n-tuples then by t+x we will denote the n-tuple (t1 + x1, . . . , tn+xn).
Theorem 3. Let µ be a measure such that [a,b] is a finite µ-measurable set, f, g and f g be µ-integrable functions on[a,b]. Let Λ = (λ1, . . . λn) be a positive n-tuple, [a,a+λ]⊂[a,b]and A be a real number such that
Aµ [a,a+ Λ]= Z
[a,b]
gdµ,
and let one of the following cases be satisfied µ-almost everywhere:
1. g(t)≤A and f(t)≥f(a+ Λ), for t∈[a,a+ Λ];
g(t)≥0 and f(t)≤f(a+ Λ),for t∈[a,b]\[a,a+ Λ];
2. g(t)≥A and f(t)≤f(a+ Λ),for t∈[a,a+ Λ];
g(t)≤0 and f(t)≥f(a+ Λ), for t∈[a,b]\[a,a+ Λ].
Then
A Z
[a,a+Λ]
fdµ≥ Z
[a,b]
f gdµ.
This is a direct multidimensional generalization of the right-hand side of Steffensen’s inequality which states that if f is a nonincreasing function on [a, b] andg is integrable withg(t)∈[0,1], then the following inequalities hold
(3)
Z b b−λ
f(t)dt≤ Z b
a
f(t)g(t)dt≤ Z a+λ
a
f(t)dt, λ= Z b
a
g(t)dt.
Generalization of the left-hand side of the inequality is given in the following theorem.
Theorem 4. Let µ be a measure such that [a,b] is a finite µ-measurable set, f, g and f g be µ-integrable functions on [a,b]. Let Λ = (λ1, . . . , λn) be a positive n-tuple, [b−Λ,b]⊂[a,b] andA be a real number such that
Aµ [b−Λ,b]= Z
[a,b]
gdµ,
and let one of the following three cases be satisfied µ-almost everywhere:
1. g(t)≤A and f(t)≤f(b−Λ), for t∈[b−Λ,b];
g(t)≥0 and f(t)≥f(b−Λ),for t∈[a,b]\[b−Λ,b];
2. g(t)≥A and f(t)≥f(b−Λ),for t∈[b−Λ,b];
g(t)≤0 and f(t)≤f(b−Λ), for t∈[a,b]\[b−Λ,b].
Then
A Z
[b−Λ,b]fdµ≤ Z
[a,b]
f gdµ.
Proofs of the above mentioned theorems are given in [7]. Now, we are in the situation to pointed out the following result which improve classical Jensen’s inequality.
Theorem5. Let f : (a, b)→Rbe a differentiable function,(x1, x2, . . . , xn)
∈ (a, b)n be a nonincreasing n-tuple, and p1, . . . , pn be a real numbers such that there exists integer k∈ {1,2, . . . , n−1} with properties
1 Pn
n
X
i=1
pixi ∈ (xk+1, xk] (4)
Pj ≥ 0, j= 1,2, . . . , k;
P¯j ≥ 0 j=k+ 1, . . . , n, Pn>0, (5)
where Pj =
j
P
i=1
pi andP¯j =
n
P
i=j
pi. If f has property that
f0(x) ≤ f0P1
n
n
P
i=1
pixi, for x≤ P1
n
n
X
i=1
pixi and
f0(x) ≥ f0P1
n
Pn i=1
pixi, for x≥ P1
n
n
X
i=1
pixi, (6)
then(1) holds.
Proof. We use idea from [5]. Settingg(t) =gjon (xj+1, xj],j= 1, . . . , n−1, where
gj = PPj
n, and g(xn) =gn= 1,
we easily check that the following holds:
gj ≤1 for j=n, . . . , k, gj ≥0 for j= 1, . . . , k and after some simple calculations we get
λ= Z x1
xn
g(t)dt= P1
n
n
X
i=1
pixi−xn.
Functions g and −f0 satisfy assumptions from Theorem 3, case 1, for n= 1, so, the following inequality holds
Z x1
xn
f0(t)g(t)dt≥
Z xn+λ xn
f0(t)dt and we get
n−1
X
k=1
f(xk)−f(xk+1)gk ≥f(xn+λ)−f(xn)
wherefrom it can be obtained (1).
Furthermore, Theorem 5 is also true if (x1, . . . , xn) is a nondecreasing se- quence; the only change needed in the hypothesis is to require that
1 Pn
n
X
i=1
pixi ∈(xk, xk+1].
In this case, the proof is based on the inequality described in Theorem 4.
Once more, we stress that under the conditions of the previous Theorem 5, the Jensen inequality holds not only for convex functions but also for the wider class of functions.
3. FURTHER RESULT
Let c∈(a, b) andλ be a real number. We will say that f : (a, b)→R has property K1(c,λ), if for anyz, y∈(a, b), z, y≥c,
(7) f(z)−f(y)
z−y ≥λ
holds.
If for anyz, y ∈(a, b), z, y ≤cinequality (7) holds then we will say that f has property K2(c,λ).
If in (7) the reversed inequality holds, then we will say that f has property RK1(c,λ) or RK2(c,λ) respectively.
If f is differentiable function on (a, b) and if f has a property K1(c,λ) for somec∈(a, b), then for f0 the following holds
f0(x)≥λ, ∀x≥c.
Obviously, if f is a nondecreasing function on (a, b) then f has properties K1(c,0) and K2(c,0) for anyc∈(a, b). Also, it is known that a convex function
f defined on an interval (a, b) has properties K1(c, f+0(c)) and RK2(c, f−0 (c)) for anyc∈(a, b).
Theorem 6. Let (x1, . . . , xn) ∈ (a, b)n be a nonincreasing sequence and p1, . . . , pn be real numbers such thatPn=
n
P
i=1
pi >0. Let us suppose that there exists an integer m∈ {1,2, . . . , n−1} such that x¯= P1
n
n
P
i=1
pixi∈(xm+1, xm].
1. If Pi ≥0,for i= 1, . . . , m,P¯i≥0,for i=m, . . . , n and iff : (a, b)→ Rhas properties K1(¯x, λ) and RK2(¯x, λ) then
(8) fP1
n
n
P
i=1
pixi
≤ P1
n
n
X
i=1
pif(xi)
holds, where Pi =
i
P
j=1
pj and P¯i=
n
P
j=i
pj.
2. If Pi ≥0,for i= 1, . . . , m, P¯i≤0,for i=m, . . . , nand if f : (a, b)→ Rhas properties K1(¯x, λ) and K2(¯x, λ) then(8) holds.
3. Similarly, if Pi ≤ 0, for i= 1, . . . , m, P¯i ≥0, for i=m, . . . , n and if f : (a, b)→R has propertiesRK1(¯x, λ) and RK2(¯x, λ) then(8) holds.
Proof. The proof is based on the following identity [4]:
f(¯x)−P1
n
n
X
i=1
pif(xi) =
m−1
X
i=1
λ(xi−xi+1)− f(xi)−f(xi+1)PPi
n
+λ(xm−x)¯ −(f(xm)−f(¯x)PPm
n
+f(¯x)−f(xm+1)−λ(¯x−xm+1)P¯m+1P
n
+
n−1
X
i=m+1
f(xi)−f(xi+1)−λ(xi−xi+1)P¯Pi+1
n ,
wherePi =
i
P
j=1
pj and ¯Pi =
n
P
j=i
pj.
Let us suppose that assumptions from case (1) hold. Then the right-hand side of the previous identity is non-positive, so the left-hand side is also non- positive, and inequality (8) is proved. Other cases can be proved on the same
way.
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Received by the editors: March 9, 1998.